Python - 如何使用 for 循环累积逗号

Question

社区-这是我的第一个post，所以如果我未能正确显示此消息，请原谅我。我正在尝试在测试用例中添加如下所示的逗号。似乎有比我在下面编码的方法更有效的方法；但是，我想使用下面的代码解决问题。我到底错过了什么？

def get_country_codes(prices):
    country_prices = prices
    p = ""
    for i in country_prices:
        if i.isalpha() and i == ",": 
            p = p + i[0] + ","
    return (p)

我的代码正在返回：

Test Failed: expected NZ, KR, DK but got 
Test Failed: expected US, AU, JP but got 
Test Failed: expected AU, NG, MX, BG, ES but got 
Test Failed: expected CA but got 

from test import testEqual

testEqual(get_country_codes("NZ0, KR00, DK"), "NZ, KR, DK")
testEqual(get_country_codes("US, AU, JP0"), "US, AU, JP")
testEqual(get_country_codes("AU, NG0, MX0, BG0, ES"), "AU, NG, MX, BG, ES")
testEqual(get_country_codes("CA"), "CA")

Answer 1

最好接受 list 而不是 string 类型作为 get_country_codes 的参数。这将使您不必担心解析字符串和忽略逗号。我推荐 Joining Lists and Splitting Strings by diveintopython.net。

此代码接受一个列表，遍历它，拆分 $ 上的每个值，获取第一个标记，并检查它是否通过 isalpha()。如果是，它会附加返回的列表。

def get_country_codes(prices):
    """Get country codes given a list of prices.

    :param prices: list of prices
    :return: list of alphabetical country codes
    """
    countries = []
    for price in prices:
        country = price.split('$')[0]
        if country.isalpha():
            countries.append(country)
    return countries

# As a proof, this line will concatenate the returned country codes
# with a comma and a space:
print(', '.join(get_country_codes(["NZ0", "KR00", "DK"])))
# As another proof, if your input has to be a string:
print(get_country_codes("US, AU, JP0".split(", ")))
print(get_country_codes(["AU", "NG0", "MX0", "BG0", "ES"]))
print(get_country_codes(["CA"]))

代码returns

NZ, KR, DK
['US', 'AU', 'JP']
['AU', 'NG', 'MX', 'BG', 'ES']
['CA']

最后，做一个断言：

testEqual(get_country_codes(["NZ0", "KR00", "DK"]), ["NZ", "KR", "DK"])

Answer 2

使用您的代码... 对于每个字符检查是否被允许。在允许的字符中包含“,”和space。

def get_country_codes(prices):
    country_prices = prices
    p = ""
    for i in country_prices:
        if i.isalpha() or i == "," or i == " ": 
            p = p + i[0]
    return (p)

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